Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $k \neq 0$. $q = \dfrac{k + 5}{-4k^3 + 144k} \div \dfrac{k + 10}{k^3 + 16k^2 + 60k} $
Dividing by an expression is the same as multiplying by its inverse. $q = \dfrac{k + 5}{-4k^3 + 144k} \times \dfrac{k^3 + 16k^2 + 60k}{k + 10} $ First factor out any common factors. $q = \dfrac{k + 5}{-4k(k^2 - 36)} \times \dfrac{k(k^2 + 16k + 60)}{k + 10} $ Then factor the quadratic expressions. $q = \dfrac {k + 5} {-4k(k + 6)(k - 6)} \times \dfrac {k(k + 6)(k + 10)} {k + 10} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac {(k + 5) \times k(k + 6)(k + 10) } { -4k(k + 6)(k - 6) \times (k + 10)} $ $q = \dfrac {k(k + 6)(k + 10)(k + 5)} {-4k(k + 6)(k - 6)(k + 10)} $ Notice that $(k + 6)$ and $(k + 10)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac {k\cancel{(k + 6)}(k + 10)(k + 5)} {-4k\cancel{(k + 6)}(k - 6)(k + 10)} $ We are dividing by $k + 6$ , so $k + 6 \neq 0$ Therefore, $k \neq -6$ $q = \dfrac {k\cancel{(k + 6)}\cancel{(k + 10)}(k + 5)} {-4k\cancel{(k + 6)}(k - 6)\cancel{(k + 10)}} $ We are dividing by $k + 10$ , so $k + 10 \neq 0$ Therefore, $k \neq -10$ $q = \dfrac {k(k + 5)} {-4k(k - 6)} $ $ q = \dfrac{-(k + 5)}{4(k - 6)}; k \neq -6; k \neq -10 $